10th Maths Quarterly Exam 2025-26 | Full Question Paper with Solutions
Original Question Paper
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PART - I (14 x 1 = 14)
Choose the best answer:
1. If there are 1024 relations from a set A = {1,2,3,4,5} to a set B. Then the number of elements in B is
Answer: b) 2
The number of relations from A to B is \( 2^{n(A) \times n(B)} \).
We are given that this is 1024.
So, \( 2^{5 \times m} = 1024 \).
We know that \( 1024 = 2^{10} \).
Therefore, \( 2^{5m} = 2^{10} \).
Equating the powers, we get \( 5m = 10 \), which gives \( m = 2 \).
Thus, the number of elements in B is 2.
2. Let A = {1, 2, 3, 4}, B = {4, 8, 9, 10}. A function f : A→B given by f = {(1, 4), (2, 8), (3, 9), (4, 10)} is a
Answer: b) one-to-one function
3. Let \( f(x) = \sqrt{1+x^2} \) then
Answer: c) \( f(xy) \le f(x).f(y) \)
LHS: \( f(xy) = \sqrt{1+(xy)^2} = \sqrt{1+x^2y^2} \).
RHS: \( f(x) \cdot f(y) = \sqrt{1+x^2} \cdot \sqrt{1+y^2} = \sqrt{(1+x^2)(1+y^2)} = \sqrt{1+x^2+y^2+x^2y^2} \).
Since \(x^2 \ge 0\) and \(y^2 \ge 0\), the term \(x^2+y^2\) is always non-negative.
Therefore, \(1+x^2+y^2+x^2y^2 \ge 1+x^2y^2\).
Taking the square root of both sides, we get \( \sqrt{1+x^2+y^2+x^2y^2} \ge \sqrt{1+x^2y^2} \).
This means \( f(x)f(y) \ge f(xy) \), which is equivalent to \( f(xy) \le f(x)f(y) \).
4. If A = \( 2^{65} \) and B = \( 2^{64} + 2^{63} + 2^{62} + \dots + 2^0 \). which of the following is true?
Answer: d) A is larger than B by 1
The sum of a GP is \( S_n = a(r^n - 1)/(r-1) \).
So, \( B = 1(2^{65} - 1)/(2-1) = 2^{65} - 1 \).
Given, \( A = 2^{65} \).
Therefore, \( A = B + 1 \), which means A is larger than B by 1.
5. The value of \( (1^3+2^3+3^3+\dots+15^3) - (1+2+3+\dots+15) \) is
Answer: c) 14280
Here, \( n = 15 \).
Let \( S = 1+2+\dots+15 = \frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 120 \).
The sum of cubes is \( S^2 = 120^2 = 14400 \).
The required value is \( S^2 - S = 14400 - 120 = 14280 \).
6. \( \frac{x}{x^2 - 25} - \frac{8}{x^2+6x+5} \) gives
Answer: c) \( \frac{x^2 - 7x + 40}{(x^2-25)(x+1)} \)
The expression becomes: \( \frac{x}{(x-5)(x+5)} - \frac{8}{(x+5)(x+1)} \).
The LCM of the denominators is \( (x-5)(x+5)(x+1) \).
\( = \frac{x(x+1) - 8(x-5)}{(x-5)(x+5)(x+1)} \)
\( = \frac{x^2+x - 8x+40}{(x^2-25)(x+1)} \)
\( = \frac{x^2 - 7x + 40}{(x^2-25)(x+1)} \)
7. The number of excluded values for the expression \( \frac{x^3 - x^2 - 10x + 8}{x^4 + 8x^2 - 9} \)
Answer: b) 2
Let \( y = x^2 \). The equation becomes \( y^2 + 8y - 9 = 0 \).
Factoring gives \( (y+9)(y-1) = 0 \).
So, \( y = -9 \) or \( y = 1 \).
Substituting back \( x^2 = y \):
Case 1: \( x^2 = -9 \). This has no real solutions.
Case 2: \( x^2 = 1 \). This gives \( x = 1 \) and \( x = -1 \).
There are 2 real excluded values.
8. If in triangle ABC and EDF, \( \frac{AB}{DE} = \frac{BC}{FD} \), then they will be similar, when
Answer: c) ∠B = ∠D
In ΔABC, the angle between sides AB and BC is ∠B.
In ΔEDF, the angle between sides DE and FD is ∠D.
Therefore, for the triangles to be similar, we must have \( \angle B = \angle D \).
9. In a given figure ST||QR, PS=2cm and SQ=3cm. Then the ratio of the area of ΔPQR to area of ΔPST is
Answer: a) 25:4
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
\( \frac{\text{Area}(\Delta PQR)}{\text{Area}(\Delta PST)} = \left(\frac{PQ}{PS}\right)^2 \)
Given, PS = 2 cm and SQ = 3 cm. So, PQ = PS + SQ = 2 + 3 = 5 cm.
Ratio of areas = \( \left(\frac{5}{2}\right)^2 = \frac{25}{4} \).
The ratio is 25:4.
10. If slope of the line PQ is \( 1/\sqrt{3} \) then slope of the perpendicular bisector of PQ is
Answer: b) \( -\sqrt{3} \)
Let the slope of PQ be \( m_1 = 1/\sqrt{3} \).
Let the slope of the perpendicular bisector be \( m_2 \).
Then, \( m_1 \times m_2 = -1 \).
\( (1/\sqrt{3}) \times m_2 = -1 \).
\( m_2 = -\sqrt{3} \).
11. The area of Quadrilateral formed by points (a, a), (-a, a), (-a, -a), (a, -a) is 64 sq.units. What is the value of a
Answer: a) 4
Area of a square = side\(^2\).
Area = \( (2a)^2 = 4a^2 \).
Given, Area = 64 sq. units.
\( 4a^2 = 64 \implies a^2 = 16 \implies a = 4 \) (assuming a > 0).
12. (2, 1) is the point of intersection of two lines
Answer: b) x+y=3; 3x+y=7
For option (b):
First line: \( x+y = 2+1 = 3 \). (Satisfied)
Second line: \( 3x+y = 3(2)+1 = 6+1 = 7 \). (Satisfied)
Since (2, 1) satisfies both equations, it is their point of intersection.
13. The value of \( \sin^2\theta + \frac{1}{1 + \tan^2\theta} \) is equal to
Answer: b) 1
The expression becomes \( \sin^2\theta + \frac{1}{\sec^2\theta} \).
Since \( \frac{1}{\sec^2\theta} = \cos^2\theta \), the expression is \( \sin^2\theta + \cos^2\theta \).
Using the Pythagorean identity \( \sin^2\theta + \cos^2\theta = 1 \).
14. If \( x = a\tan\theta \) and \( y = b\sec\theta \) then
Answer: a) \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \)
We use the identity \( \sec^2\theta - \tan^2\theta = 1 \).
Substituting the expressions for \( \sec\theta \) and \( \tan\theta \):
\( \left(\frac{y}{b}\right)^2 - \left(\frac{x}{a}\right)^2 = 1 \)
\( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \)
PART - II (10 x 2 = 20)
Answer any 10 questions. Question No.28 is compulsory.
15. Define identity function.
An identity function is a function that maps every element of a set to itself. For a set A, the identity function \( f: A \rightarrow A \) is defined by \( f(x) = x \) for all \( x \in A \).
16. If A = {1, 3, 5} B = {2, 3} then (i) find AxB and BxA.
Given, A = {1, 3, 5} and B = {2, 3}.
(i) A x B (Cartesian Product):
\( A \times B = \{(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)\} \)
(ii) B x A (Cartesian Product):
\( B \times A = \{(2, 1), (2, 3), (2, 5), (3, 1), (3, 3), (3, 5)\} \)
17. Find K if f o f(k) = 5 where f(k) = 2k-1.
Given \( f(k) = 2k-1 \).
\( f \circ f(k) = f(f(k)) = f(2k-1) \)
\( = 2(2k-1) - 1 \)
\( = 4k - 2 - 1 = 4k - 3 \)
We are given \( f \circ f(k) = 5 \).
So, \( 4k - 3 = 5 \)
\( 4k = 8 \)
\( k = 2 \)
18. Find the least positive value of x such that 5x ≡ 4 (mod 6).
The congruence \( 5x \equiv 4 \pmod{6} \) means that \( 5x - 4 \) is a multiple of 6.
So, \( 5x - 4 = 6n \) for some integer \( n \).
We need to find the least positive integer \( x \) that satisfies this.
Let's test positive values for x:
If \( x=1 \), \( 5(1) - 4 = 1 \), which is not a multiple of 6.
If \( x=2 \), \( 5(2) - 4 = 10 - 4 = 6 \), which is a multiple of 6.
Therefore, the least positive value of x is 2.
19. If \( 1+2+3+\dots+n = 666 \) then find n.
The sum of the first n natural numbers is given by the formula \( \frac{n(n+1)}{2} \).
Given, \( \frac{n(n+1)}{2} = 666 \).
\( n(n+1) = 666 \times 2 = 1332 \)
\( n^2 + n - 1332 = 0 \)
We need to find two consecutive integers whose product is 1332.
We can factorize 1332: \( 1332 = 2 \times 666 = 2 \times 2 \times 333 = 4 \times 9 \times 37 = 36 \times 37 \).
So, \( n(n+1) = 36 \times 37 \).
Therefore, \( n = 36 \).
20. Simplify: \( \frac{x}{x-y} + \frac{y}{y-x} \)
We can rewrite the second term by factoring out -1 from the denominator:
\( \frac{y}{y-x} = \frac{y}{-(x-y)} = -\frac{y}{x-y} \)
Now the expression becomes:
\( \frac{x}{x-y} - \frac{y}{x-y} \)
Since the denominators are the same, we can combine the numerators:
\( \frac{x-y}{x-y} = 1 \) (provided \( x \ne y \)).
21. Find the sum and and product of the roots of the quadratic equation \( 3y^2-y-4=0 \).
For a quadratic equation \( ax^2+bx+c=0 \),
Sum of roots \( (\alpha + \beta) = -b/a \)
Product of roots \( (\alpha\beta) = c/a \)
In the given equation \( 3y^2-y-4=0 \), we have \( a=3, b=-1, c=-4 \).
Sum of roots = \( -(-1)/3 = 1/3 \).
Product of roots = \( -4/3 \).
22. If the difference between a number and its reciprocal is 24/5, find the number.
Let the number be \( x \). Its reciprocal is \( 1/x \).
Given, \( x - \frac{1}{x} = \frac{24}{5} \).
Multiplying the entire equation by \( 5x \) to clear fractions:
\( 5x^2 - 5 = 24x \)
\( 5x^2 - 24x - 5 = 0 \)
Factoring the quadratic equation:
\( 5x^2 - 25x + x - 5 = 0 \)
\( 5x(x - 5) + 1(x - 5) = 0 \)
\( (5x + 1)(x - 5) = 0 \)
The possible values for the number are \( x = 5 \) or \( x = -1/5 \).
23. In the figure DE||AC and DC||AP, Prove that \( \frac{BE}{EC} = \frac{BC}{CP} \).
Proof:
Step 1: In ΔABC, we are given that DE || AC.
By the Basic Proportionality Theorem (Thales' Theorem), if a line is parallel to one side of a triangle, it divides the other two sides proportionally.
Therefore, \( \frac{BE}{EC} = \frac{BD}{DA} \) --- (1)
Step 2: In ΔABP, we are given that DC || AP.
Again, by the Basic Proportionality Theorem,
\( \frac{BC}{CP} = \frac{BD}{DA} \) --- (2)
Step 3: From equations (1) and (2), we can see that both ratios are equal to \( \frac{BD}{DA} \).
Therefore, we can equate them:
\( \frac{BE}{EC} = \frac{BC}{CP} \)
Hence Proved.
24. Find the value of 'a' for which the given points are collinear (2, 3) (4, a) and (6, -3).
If three points are collinear, the area of the triangle formed by them is 0.
The formula for the area of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is:
Area \( = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \)
Setting the area to 0:
\( \frac{1}{2} |2(a - (-3)) + 4(-3 - 3) + 6(3 - a)| = 0 \)
\( |2(a + 3) + 4(-6) + 6(3 - a)| = 0 \)
\( |2a + 6 - 24 + 18 - 6a| = 0 \)
\( |-4a| = 0 \)
\( 4a = 0 \implies a = 0 \)
25. A cat is located at the point (-6, -4) in xy plane. A bottle of milk is kept at (5, 11). The cat wish to consume the milk travelling through shortest possible distance. Find the equation of the path it needs to take.
The shortest path between two points is a straight line. We need to find the equation of the line passing through A(-6, -4) and B(5, 11).
First, find the slope (m):
\( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{11 - (-4)}{5 - (-6)} = \frac{15}{11} \)
Now, use the point-slope form of the equation: \( y - y_1 = m(x - x_1) \)
Using point A(-6, -4):
\( y - (-4) = \frac{15}{11}(x - (-6)) \)
\( y + 4 = \frac{15}{11}(x + 6) \)
\( 11(y + 4) = 15(x + 6) \)
\( 11y + 44 = 15x + 90 \)
\( 15x - 11y + 90 - 44 = 0 \)
\( 15x - 11y + 46 = 0 \)
This is the equation of the path.
26. Find the equation of a straight line which is parallel to the line 3x-7y=12 and passing through the point (6, 4).
The equation of a line parallel to \( ax + by + c = 0 \) is of the form \( ax + by + k = 0 \).
So, the equation of the line parallel to \( 3x - 7y = 12 \) is \( 3x - 7y + k = 0 \).
This line passes through the point (6, 4). We substitute these values to find k:
\( 3(6) - 7(4) + k = 0 \)
\( 18 - 28 + k = 0 \)
\( -10 + k = 0 \implies k = 10 \)
The required equation is \( 3x - 7y + 10 = 0 \).
27. Prove that \( \sqrt{\frac{1 + \sin\theta}{1 - \sin\theta}} = \sec\theta + \tan\theta \). (Assuming the standard form of this question)
LHS = \( \sqrt{\frac{1 + \sin\theta}{1 - \sin\theta}} \)
Multiply the numerator and denominator inside the square root by the conjugate of the denominator, which is \( (1 + \sin\theta) \):
LHS = \( \sqrt{\frac{(1 + \sin\theta)(1 + \sin\theta)}{(1 - \sin\theta)(1 + \sin\theta)}} \)
= \( \sqrt{\frac{(1 + \sin\theta)^2}{1 - \sin^2\theta}} \)
Using the identity \( \cos^2\theta = 1 - \sin^2\theta \):
= \( \sqrt{\frac{(1 + \sin\theta)^2}{\cos^2\theta}} \)
= \( \frac{1 + \sin\theta}{\cos\theta} \)
= \( \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta} \)
= \( \sec\theta + \tan\theta \) = RHS
Hence Proved.
28. (Compulsory) If in an A.P, a = -10, d = 2, how many terms are needed to reach the term 0?
Let the A.P be \( T_1, T_2, T_3, \dots \)
Given the first term \( a = -10 \) and the common difference \( d = 2 \).
We want to find which term, \( T_n \), is equal to 0.
The formula for the n-th term of an A.P is \( T_n = a + (n-1)d \).
Set \( T_n = 0 \):
\( 0 = -10 + (n-1)2 \)
\( 10 = (n-1)2 \)
\( 5 = n-1 \)
\( n = 6 \)
So, the 6th term of the A.P is 0. 6 terms are needed to reach the term 0.
PART - III (10 x 5 = 50)
Answer any 10 questions. Q.No.42 is compulsory.
29. Let A = The set of all natural numbers less than 8, B=The set of all prime numbers less than 8, C=The set of even prime number. Verify that (A∩B) x C = (AxC) ∩ (BxC).
First, let's write the sets in roster form:
A = {1, 2, 3, 4, 5, 6, 7}
B = {2, 3, 5, 7}
C = {2} (2 is the only even prime number)
LHS: (A∩B) x C
A∩B = {2, 3, 5, 7}
(A∩B) x C = {2, 3, 5, 7} x {2} = {(2, 2), (3, 2), (5, 2), (7, 2)} --- (1)
RHS: (AxC) ∩ (BxC)
A x C = {1, 2, 3, 4, 5, 6, 7} x {2} = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (7, 2)}
B x C = {2, 3, 5, 7} x {2} = {(2, 2), (3, 2), (5, 2), (7, 2)}
(AxC) ∩ (BxC) = {(2, 2), (3, 2), (5, 2), (7, 2)} --- (2)
From (1) and (2), we see that LHS = RHS. Hence verified.
30. A function f: [-5, 9] → R is defined as follows:
\( f(x) = \begin{cases} 6x + 1 & -5 \le x < 2 \\ 5x^2 - 1 & 2 \le x < 6 \\ 3x - 4 & 6 \le x \le 9 \end{cases} \).
Find
i) f(-3)+f(2)
ii) f(7)-f(1)
iii) 2f(4)+f(8)
iv) \( \frac{2f(-2)-f(6)}{f(4)+f(-2)} \)
i) f(-3) + f(2)
f(-3) is in the interval \( -5 \le x < 2 \), so we use \( 6x+1 \). f(-3) = 6(-3) + 1 = -17.
f(2) is in the interval \( 2 \le x < 6 \), so we use \( 5x^2-1 \). f(2) = 5(2)² - 1 = 19.
f(-3) + f(2) = -17 + 19 = 2.
ii) f(7) - f(1)
f(7) is in the interval \( 6 \le x \le 9 \), so we use \( 3x-4 \). f(7) = 3(7) - 4 = 17.
f(1) is in the interval \( -5 \le x < 2 \), so we use \( 6x+1 \). f(1) = 6(1) + 1 = 7.
f(7) - f(1) = 17 - 7 = 10.
iii) 2f(4) + f(8)
f(4) is in the interval \( 2 \le x < 6 \), so we use \( 5x^2-1 \). f(4) = 5(4)² - 1 = 79.
f(8) is in the interval \( 6 \le x \le 9 \), so we use \( 3x-4 \). f(8) = 3(8) - 4 = 20.
2f(4) + f(8) = 2(79) + 20 = 158 + 20 = 178.
iv) \( \frac{2f(-2)-f(6)}{f(4)+f(-2)} \)
f(-2) is in \( -5 \le x < 2 \), so f(-2) = 6(-2) + 1 = -11.
f(6) is in \( 6 \le x \le 9 \), so f(6) = 3(6) - 4 = 14.
f(4) = 79 (from part iii).
The expression is \( \frac{2(-11) - 14}{79 + (-11)} = \frac{-22 - 14}{68} = \frac{-36}{68} = -\frac{9}{17} \).
31. If f(x) = 2x+3, g(x) = 1-2x and h(x)=3x. Prove that fo(goh)=(fog)oh.
To prove the associative property of function composition, we will evaluate the Left Hand Side (LHS) and the Right Hand Side (RHS) separately.
LHS = f o (g o h)
First, find g o h(x):
\( g \circ h(x) = g(h(x)) = g(3x) = 1 - 2(3x) = 1 - 6x \)
Now, find f o (g o h)(x):
\( f \circ (g \circ h)(x) = f(g \circ h(x)) = f(1 - 6x) \)
\( = 2(1 - 6x) + 3 \)
\( = 2 - 12x + 3 \)
\( = 5 - 12x \) --- (1)
RHS = (f o g) o h
First, find f o g(x):
\( f \circ g(x) = f(g(x)) = f(1 - 2x) = 2(1 - 2x) + 3 \)
\( = 2 - 4x + 3 \)
\( = 5 - 4x \)
Now, find (f o g) o h(x):
\( (f \circ g) \circ h(x) = (f \circ g)(h(x)) = (f \circ g)(3x) \)
\( = 5 - 4(3x) \)
\( = 5 - 12x \) --- (2)
From (1) and (2), we see that LHS = RHS. Hence, \( f \circ (g \circ h) = (f \circ g) \circ h \) is proved.
32. Find the HCF of 396, 504, 636.
We will use Euclid's Division Algorithm to find the HCF.
Step 1: Find HCF of 504 and 396.
\( 504 = 396 \times 1 + 108 \)
\( 396 = 108 \times 3 + 72 \)
\( 108 = 72 \times 1 + 36 \)
\( 72 = 36 \times 2 + 0 \)
The remainder is 0. So, the HCF of 504 and 396 is 36.
Step 2: Find HCF of 636 and the result from Step 1 (which is 36).
\( 636 = 36 \times 17 + 24 \)
\( 36 = 24 \times 1 + 12 \)
\( 24 = 12 \times 2 + 0 \)
The remainder is 0. So, the HCF of 636 and 36 is 12.
Therefore, the HCF of 396, 504, and 636 is 12.
33. The sum of first n, 2n and 3n terms of an A.P. are S₁, S₂, and S₃ respectively. Prove that S₃ = 3(S₂ - S₁).
Let the first term of the A.P be 'a' and the common difference be 'd'. The formula for the sum of the first 'k' terms of an A.P is \( S_k = \frac{k}{2}[2a + (k-1)d] \).
Using this formula, we can write the expressions for S₁, S₂, and S₃:
\( S_1 = \frac{n}{2}[2a + (n-1)d] \) --- (1)
\( S_2 = \frac{2n}{2}[2a + (2n-1)d] = n[2a + (2n-1)d] \) --- (2)
\( S_3 = \frac{3n}{2}[2a + (3n-1)d] \) --- (3)
Now, let's evaluate the RHS of the equation we need to prove: 3(S₂ - S₁)
First, calculate S₂ - S₁:
\( S_2 - S_1 = n[2a + (2n-1)d] - \frac{n}{2}[2a + (n-1)d] \)
Take \( \frac{n}{2} \) as a common factor:
\( = \frac{n}{2} \left( 2[2a + (2n-1)d] - [2a + (n-1)d] \right) \)
\( = \frac{n}{2} [ (4a + (4n-2)d) - (2a + (n-1)d) ] \)
\( = \frac{n}{2} [ 4a - 2a + (4n-2)d - (n-1)d ] \)
\( = \frac{n}{2} [ 2a + (4n-2 - n+1)d ] \)
\( = \frac{n}{2} [ 2a + (3n-1)d ] \)
Now, multiply by 3:
\( 3(S_2 - S_1) = 3 \times \frac{n}{2} [ 2a + (3n-1)d ] \)
\( = \frac{3n}{2} [ 2a + (3n-1)d ] \)
This is exactly the expression for S₃ from equation (3).
Thus, LHS (S₃) = RHS (3(S₂ - S₁)). Hence proved.
34. Find the values of m and n if the polynomial is a perfect square: 36x⁴ - 60x³ + 61x² - mx + n.
We use the long division method to find the square root of the polynomial. For it to be a perfect square, the remainder must be zero.
6x² - 5x + 3
_________________________
6x² | 36x⁴ - 60x³ + 61x² - mx + n
|-(36x⁴)
|_________________________
12x²-5x | - 60x³ + 61x²
| -(- 60x³ + 25x²)
| ___________________
12x²-10x+3| 36x² - mx + n
| -(36x² - 30x + 9)
| ________________
| (-m+30)x + (n-9)
Explanation of Steps:
- The square root of \(36x^4\) is \(6x^2\). Write this in the quotient and divisor. Subtract \( (6x^2)^2 = 36x^4 \).
- Bring down the next two terms: \(-60x^3 + 61x^2\). Double the current quotient to get the new divisor's first part: \(2 \times 6x^2 = 12x^2\).
- Divide the first term of the new dividend (\(-60x^3\)) by the first term of the new divisor (\(12x^2\)) to get \(-5x\). This is the next term in the quotient and divisor.
- Multiply \(-5x\) by the new divisor \( (12x^2 - 5x) \) to get \(-60x^3 + 25x^2\). Subtract this.
- Bring down the last two terms: \(-mx + n\). The new dividend is \(36x^2 - mx + n\). Double the current quotient \((6x^2 - 5x)\) to get \(12x^2 - 10x\).
- Divide \(36x^2\) by \(12x^2\) to get \(+3\). This is the last term in the quotient and divisor.
- Multiply \(+3\) by the new divisor \((12x^2 - 10x + 3)\) to get \(36x^2 - 30x + 9\). Subtract this.
The remainder is \( (-m+30)x + (n-9) \).
Since the polynomial is a perfect square, the remainder must be 0.
\( (-m+30)x + (n-9) = 0x + 0 \)
Equating the coefficients of x and the constant terms to zero:
\( -m + 30 = 0 \implies m = 30 \)
\( n - 9 = 0 \implies n = 9 \)
Therefore, the values are m = 30 and n = 9.
35. Solve: pq x² - (p+q)² x + (p+q)² = 0.
This is a quadratic equation in the form \( Ax^2 + Bx + C = 0 \), where:
A = pq
B = \( -(p+q)^2 \)
C = \( (p+q)^2 \)
We use the quadratic formula to solve for x: \( x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \)
Substitute the values of A, B, and C:
\( x = \frac{-(-(p+q)^2) \pm \sqrt{(-(p+q)^2)^2 - 4(pq)((p+q)^2)}}{2(pq)} \)
\( x = \frac{(p+q)^2 \pm \sqrt{(p+q)^4 - 4pq(p+q)^2}}{2pq} \)
Factor out \( (p+q)^2 \) from inside the square root:
\( x = \frac{(p+q)^2 \pm \sqrt{(p+q)^2 [(p+q)^2 - 4pq]}}{2pq} \)
Simplify the term inside the brackets: \( (p+q)^2 - 4pq = p^2 + 2pq + q^2 - 4pq = p^2 - 2pq + q^2 = (p-q)^2 \)
\( x = \frac{(p+q)^2 \pm \sqrt{(p+q)^2 (p-q)^2}}{2pq} \)
\( x = \frac{(p+q)^2 \pm (p+q)(p-q)}{2pq} \)
Now we find the two possible solutions for x:
Case 1 (with '+'):
\( x = \frac{(p+q)^2 + (p+q)(p-q)}{2pq} = \frac{(p+q)[(p+q) + (p-q)]}{2pq} \)
\( x = \frac{(p+q)[2p]}{2pq} = \frac{p+q}{q} \)
Case 2 (with '-'):
\( x = \frac{(p+q)^2 - (p+q)(p-q)}{2pq} = \frac{(p+q)[(p+q) - (p-q)]}{2pq} \)
\( x = \frac{(p+q)[p+q-p+q]}{2pq} = \frac{(p+q)[2q]}{2pq} = \frac{p+q}{p} \)
The solutions are \( x = \frac{p+q}{p} \) and \( x = \frac{p+q}{q} \).
36. State and prove Basic proportionality theorem.
Statement (Thales' Theorem): If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Given: In ΔABC, a line DE is parallel to BC (DE || BC), intersecting AB at D and AC at E.
To Prove: \( \frac{AD}{DB} = \frac{AE}{EC} \)
Construction: Join BE and CD. Draw DM ⊥ AC and EN ⊥ AB.
Proof:
In ∆ABC, D is a point on AB and E is a point on AC.
To prove: \(\frac{AD}{DB} = \frac{AE}{EC}\)
Construction: Draw a line DE || BC
| No. | Statement | Reason |
|---|---|---|
| 1. | ∠ABC = ∠ADE = ∠1 | Corresponding angles are equal because DE || BC |
| 2. | ∠ACB = ∠AED = ∠2 | Corresponding angles are equal because DE || BC |
| 3. | ∠DAE = ∠BAC = ∠3 | Both triangles have a common angle |
| ∆ABC ~ ∆ADE | By AAA similarity | |
| \(\frac{AB}{AD} = \frac{AC}{AE}\) | Corresponding sides are proportional | |
| \(\frac{AD+DB}{AD} = \frac{AE+EC}{AE}\) | Split AB and AC using the points D and E. | |
| \(1 + \frac{DB}{AD} = 1 + \frac{EC}{AE}\) | On simplification | |
| \(\frac{DB}{AD} = \frac{EC}{AE}\) | Cancelling 1 on both sides | |
| \(\frac{AD}{DB} = \frac{AE}{EC}\) | Taking reciprocals. Hence proved. |
37. Find the area of the quadrilateral whose vertices are at (-9, -2), (-8, -4), (2, 2) and (1, -3).
To find the area of the quadrilateral, we use the shoelace formula. First, let's arrange the vertices in counter-clockwise order to ensure a positive area. A rough plot shows the order A(-9, -2), B(-8, -4), D(1, -3), C(2, 2) is counter-clockwise.
Let the vertices be:
\( (x_1, y_1) = (-9, -2) \)
\( (x_2, y_2) = (-8, -4) \)
\( (x_3, y_3) = (1, -3) \)
\( (x_4, y_4) = (2, 2) \)
The formula for the area is: \( \text{Area} = \frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)| \)
Let's calculate the two parts:
Part 1: \( (x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) \)
\( = (-9)(-4) + (-8)(-3) + (1)(2) + (2)(-2) \)
\( = 36 + 24 + 2 - 4 = 58 \)
Part 2: \( (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \)
\( = (-2)(-8) + (-4)(1) + (-3)(2) + (2)(-9) \)
\( = 16 - 4 - 6 - 18 = -12 \)
Now, substitute these values back into the area formula:
\( \text{Area} = \frac{1}{2} |58 - (-12)| \)
\( = \frac{1}{2} |58 + 12| \)
\( = \frac{1}{2} |70| = 35 \)
The area of the quadrilateral is 35 square units.
38. Find the equation of a straight line through the point of intersection of the lines 8x+3y=18, 4x+5y=9 and bisecting the line segment joining the points (5, -4) and (-7, 6).
Step 1: Find the point of intersection of the two lines.
Line 1: \( 8x + 3y = 18 \) --- (1)
Line 2: \( 4x + 5y = 9 \) --- (2)
Multiply equation (2) by 2: \( 8x + 10y = 18 \) --- (3)
Subtract equation (1) from (3):
\( (8x + 10y) - (8x + 3y) = 18 - 18 \)
\( 7y = 0 \implies y = 0 \)
Substitute \(y = 0\) into equation (2):
\( 4x + 5(0) = 9 \implies 4x = 9 \implies x = 9/4 \)
So, the point of intersection is P(9/4, 0).
Step 2: Find the midpoint of the line segment joining (5, -4) and (-7, 6).
Midpoint M = \( \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \)
M = \( \left( \frac{5+(-7)}{2}, \frac{-4+6}{2} \right) = \left( \frac{-2}{2}, \frac{2}{2} \right) = (-1, 1) \)
Step 3: Find the equation of the line passing through P(9/4, 0) and M(-1, 1).
Using the two-point form \( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1) \):
Slope (m) = \( \frac{1 - 0}{-1 - 9/4} = \frac{1}{-13/4} = -\frac{4}{13} \)
Using point M(-1, 1) and the slope:
\( y - 1 = -\frac{4}{13}(x - (-1)) \)
\( y - 1 = -\frac{4}{13}(x + 1) \)
\( 13(y - 1) = -4(x + 1) \)
\( 13y - 13 = -4x - 4 \)
\( 4x + 13y - 13 + 4 = 0 \)
\( 4x + 13y - 9 = 0 \)
The required equation of the straight line is 4x + 13y - 9 = 0.
39. A mobile phone is put to use when the battery power is 100%. The percent of battery power 'y' (in decimal) remaining after using the mobile phone for x hours is assumed as y = -0.25x+1.
(i) Find the number of hours elapsed if the battery power is 40%.
(ii) How much time does it take so that the battery has no power?
The given relation is \( y = -0.25x + 1 \), where 'y' is the battery power in decimal and 'x' is the time in hours.
(i) Find x when battery power is 40%.
40% battery power means \( y = 0.40 \).
Substitute this value into the equation:
\( 0.40 = -0.25x + 1 \)
\( 0.25x = 1 - 0.40 \)
\( 0.25x = 0.60 \)
\( x = \frac{0.60}{0.25} = \frac{60}{25} = \frac{12}{5} = 2.4 \)
So, 2.4 hours have elapsed.
(ii) Find x when the battery has no power.
No power means the battery is at 0%, so \( y = 0 \).
Substitute this value into the equation:
\( 0 = -0.25x + 1 \)
\( 0.25x = 1 \)
\( x = \frac{1}{0.25} = 4 \)
It takes 4 hours for the battery to have no power.
40. A man joined a company as Assistant Manager. The company gave him a starting salary of ₹60,000 and agreed to increase his salary 5% annually. What will be his salary after 5 years?
This is a problem of geometric progression (GP).
The starting salary (first term, a) = ₹60,000.
The annual increase is 5%. This means each year the salary becomes 105% of the previous year's salary.
The common ratio (r) = \( 1 + \frac{5}{100} = 1 + 0.05 = 1.05 \).
"Salary after 5 years" refers to the salary at the beginning of the 6th year. We need to find the 6th term (\( T_6 \)) of the GP.
The formula for the n-th term of a GP is \( T_n = ar^{n-1} \).
For the 6th year (\(n=6\)):
\( T_6 = 60000 \times (1.05)^{6-1} \)
\( T_6 = 60000 \times (1.05)^5 \)
Now, let's calculate \( (1.05)^5 \):
\( (1.05)^2 = 1.1025 \)
\( (1.05)^4 = (1.1025)^2 = 1.21550625 \)
\( (1.05)^5 = 1.21550625 \times 1.05 = 1.2762815625 \)
Salary after 5 years = \( 60000 \times 1.2762815625 \)
= 76576.89375
Rounding his salary after 5 years will be ₹76,577.
41. Prove that sin²A cos²B + cos²A sin²B + cos²A cos²B + sin²A sin²B = 1.
We start with the Left Hand Side (LHS) of the equation:
LHS = \( \sin^2A \cos^2B + \cos^2A \sin^2B + \cos^2A \cos^2B + \sin^2A \sin^2B \)
Let's rearrange and group the terms with common factors:
Group 1 (terms with \( \sin^2A \)):
\( (\sin^2A \cos^2B + \sin^2A \sin^2B) \)
Factor out \( \sin^2A \):
\( = \sin^2A (\cos^2B + \sin^2B) \)
Using the identity \( \cos^2\theta + \sin^2\theta = 1 \), this simplifies to:
\( = \sin^2A (1) = \sin^2A \)
Group 2 (terms with \( \cos^2A \)):
\( (\cos^2A \sin^2B + \cos^2A \cos^2B) \)
Factor out \( \cos^2A \):
\( = \cos^2A (\sin^2B + \cos^2B) \)
Using the same identity, this simplifies to:
\( = \cos^2A (1) = \cos^2A \)
Now, add the results of the two groups:
LHS = (Result of Group 1) + (Result of Group 2)
LHS = \( \sin^2A + \cos^2A \)
Using the fundamental Pythagorean identity \( \sin^2A + \cos^2A = 1 \).
Therefore, LHS = 1.
Since LHS = 1 and RHS = 1, the identity is proved.
42. (Compulsory) If \( f(x) = x^2 - 1 \), \( g(x) = x^3 - 1 \) verify that \( f(x) \times g(x) = \text{LCM}(f(x), g(x)) \times \text{GCD}(f(x), g(x)) \).
Step 1: Factorize f(x) and g(x)
\( f(x) = x^2 - 1 = (x-1)(x+1) \)
\( g(x) = x^3 - 1 = (x-1)(x^2+x+1) \)
Step 2: Find GCD (Greatest Common Divisor)
The common factor is \( (x-1) \).
\( \text{GCD}(f(x), g(x)) = (x-1) \)
Step 3: Find LCM (Least Common Multiple)
The LCM is the product of the GCD and the remaining uncommon factors.
\( \text{LCM}(f(x), g(x)) = (x-1)(x+1)(x^2+x+1) \)
Step 4: Verify the property
LHS: \( f(x) \times g(x) = (x^2-1)(x^3-1) = (x-1)(x+1)(x-1)(x^2+x+1) \)
RHS: \( \text{LCM} \times \text{GCD} = [(x-1)(x+1)(x^2+x+1)] \times [(x-1)] \)
\( = (x-1)(x+1)(x-1)(x^2+x+1) \)
Since LHS = RHS, the property is verified.
PART - IV (2 x 8 = 16)
Answer all of the following.
43. a) Construct a triangle similar to a given triangle PQR with its sides equal to 7/3 of the corresponding sides of the triangle PQR (scale factor 7/3 > 1).
Steps of Construction:
Since the scale factor is \(\frac{7}{3}\), which is greater than 1, the new triangle will be larger than the original triangle PQR.
- Draw a triangle PQR with any suitable measurements.
- Draw a ray QX making an acute angle with QR, on the side opposite to vertex P.
- Locate 7 points \(Q_1, Q_2, Q_3, Q_4, Q_5, Q_6, Q_7\) on the ray QX such that the distances between them are equal (\(QQ_1 = Q_1Q_2 = \dots = Q_6Q_7\)).
- Join \(Q_3\) (the 3rd point, as 3 is the denominator) to R.
- Draw a line through \(Q_7\) parallel to \(Q_3R\). This line will intersect the extended line segment QR at a point R'.
- Draw a line through R' parallel to PR. This line will intersect the extended line segment QP at a point P'.
- \(\triangle P'QR'\) is the required similar triangle, with each side being \(\frac{7}{3}\) times the corresponding side of \(\triangle PQR\).
43. b) (OR) Draw a triangle ABC of base BC=8cm, ∠A=60° and the bisector of ∠A meets BC at D such that BD=6cm.
Steps of Construction:
44. a) A company initially started with 40 worker to complete the work by 150 days. Later it decided to fastern up the work increasing the number of workers as shown below.
| Number of workers (x) | 40 | 50 | 60 | 75 |
|---|---|---|---|---|
| Number of days (y) | 150 | 120 | 100 | 80 |
(ii) From the graph find the number of days required to complete the work if the company decides to opt for 120 worker?
(iii) If the work has to be completed by 200 days how many workers are required?
44. b) (OR) A garment shop announces a flat 50% discount on every purchase of items for their customers. Draw the graph for the relation between the marked price and the discount. Hence find
(i) the marked price when a customer gets a discount of ₹3250 (from graph)
(ii) the discount when the marked price is ₹2500.
